# Thread: Finding the equation for a plane

1. ## Finding the equation for a plane

The plane that contains the points (8,3,1), (2,6,3), and (4,6,2) has an equation of the form ax+by+cz=d. Find coefficients fr this equation, trying two different approaches to the problem. One method uses vectors, another does not.

2. Since the points $P_{1}, \;\ P_{2}, \;\ P_{3}$ lie in the plane, the vectors $\overline{P_{1}P_{2}}=(-6,3,2)$ and
$\overline{P_{1}P_{3}}=(-4,3,1)$ are parallel to the plane.

The cross product:

$P_{1}P_{2}\times{P_{1}P_{3}}=\begin{vmatrix}i&j&k\ \-6&3&-2\\-4&3&1\end{vmatrix}=-3i-2j-6k$
is normal to the plane.

Use this normal and the point (8,3,1):

$-3(x-8)-2(y-3)-6(z-1)=0$

And we get:

$-3x-2y-6z+36=0$

$\boxed{3x+2y+6z=36}$

Now, try the other method.